How do you convert $2=(9x+y)^2-x$ into polar form?

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Answer 1 · 2018-03-30T10:03:37

$r^2(9costheta+sintheta)^2-rcostheta-2=0$

Given:

$2=(9x+y)^2-x$

$x=rcostheta$

$y=rsintheta$

$2=(9rcostheta+rsintheta)^2-rcostheta$

$2=r^2(9costheta+sintheta)^2-rcostheta$

$r^2(9costheta+sintheta)^2-rcostheta-2=0$

How do you convert 2=(9x+y)^2-x2=(9x+y)^2-x into polar form?